Question

A ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be:

• A. $\frac{m\left(M+2m\right)}{M}{\omega }^2{R}^2$
• B. $\frac{Mm}{\left(M+m\right)}{\omega }^2{R}^2$
• C. $\frac{Mm}{\left(M+2m\right)}{\omega }^2{R}^2$
• D. $\frac{\left(M+m\right)M}{\left(M+2m\right)}{\omega }^2{R}^2$

Answer 1

Answer: (C)

$Kineticenerg{y}_{\left(rotational\right)}{K}_R=\frac{1}{2}I{\omega }^2$
$Kineticenerg{y}_{\left(rotational\right)}{K}_r=\frac{1}{2}M{V}^2$
(v=$R\omega$)
$M.I{.}_{\left(initia\right)}{I}_{ring}=M{R}^2;{\omega }_{initial}=\omega$
$M.I{.}_{\left(new\right)}{I}_{\left(system\right)}^{\prime }=M{R}^2+2m{R}^2$
${\omega }_{\left(system\right)}^{\prime }=\frac{M\omega }{M+2m}$
Solving we get loss in $K.E.$
$=\frac{Mm}{\left(M+2m\right)}{\omega }^2{R}^2$